Question: Simplify; express your answer in exponential form. Assume $k\neq 0, z\neq 0$. $\dfrac{{(k^{3})^{5}}}{{(k^{3}z^{-1})^{5}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(k^{3})^{5} = k^{15}}$ In the denominator, we can use the distributive property of exponents. ${(k^{3}z^{-1})^{5} = (k^{3})^{5}(z^{-1})^{5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{3})^{5}}}{{(k^{3}z^{-1})^{5}}} = \dfrac{{k^{15}}}{{k^{15}z^{-5}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{15}}}{{k^{15}z^{-5}}} = \dfrac{{k^{15}}}{{k^{15}}} \cdot \dfrac{{1}}{{z^{-5}}} = k^{{15} - {15}} \cdot z^{- {(-5)}} = z^{5}$.